# Group SL(2,C)

1. Sep 2, 2010

### LAHLH

Hi,

I'm just reading about the group SL(2,C). In the book that I'm using(Jones, groups reps and physics), he defines a 2x2 matrix from a generic 4 vector $$v_{\mu}$$ and a vector $$\sigma_{\mu}:=(1,\vec{\sigma})$$, as $$V:=v_{\mu}\sigma^{\mu}$$

He nows wants to invert this equation to solve for $$v_{\mu}$$, and he suggests tracing with another vector of matrices defined as $$\tilde{\sigma_{\mu}}:=(1,-\vec{\sigma})$$, and he obtains $$v_{\mu}=\tfrac{1}{2}Tr(\tilde{\sigma_{\mu}}V)$$

I can't seem to get this, starting with $$V:=v_{\mu}\sigma^{\mu}$$ and then multiplying by $$\tilde{\sigma_{\nu}}$$, leads to $$\tilde{\sigma_{\nu}}V:=v_{\mu}\sigma^{\mu}\tilde{\sigma_{\nu}}$$

Now I'm not sure what indices I'm supposed to trace with?

2. Sep 2, 2010

### chrispb

Re: Sl(2,c)

Your V is a 2x2 matrix. If you want to recover $$v_{\nu}$$ (i.e. the nu'th component of the vector v), you should take $$\frac{1}{2} Tr(\tilde{\sigma}_{\nu}V)$$, where nu is not summed over. In other words, if you want the first component, take the 2x2 matrix $$\tilde{\sigma}_1$$, left-multiply (though inside a trace order doesn't matter) it with the 2x2 matrix V, and take half the trace of the resultant 2x2 matrix. Try a simple example if this isn't clear; pick (1,0,0,0) and try to recover v_0.

3. Sep 2, 2010

### Fredrik

Staff Emeritus
Re: Sl(2,c)

I'm going to write all indices as subscripts. $\{I,\sigma_1,\sigma_2,\sigma_3\}$, is a basis for the real vector space of complex 2×2 self-adjoint matrices. I'll call that space V. If we define an inner product by

$$\langle A,B\rangle=\frac{1}{2}\operatorname{Tr}(A^\dagger B)$$

for all A,B in V, it's an orthonormal basis. So if we define $\sigma_0=I$, any $x \in V$ can be expressed as $x=x_\mu \sigma_\mu$, with $x_\mu=\langle\sigma_\mu,x\rangle$. Note that all the $x_\mu$ are real. (This is implied by the facts that V is a real vector space and that $\{\sigma_0,\sigma_1,\sigma_2,\sigma_3\}$ is a basis of V).

$$x_\mu=\langle\sigma_\mu,x\rangle=\frac{1}{2}\operatorname{Tr}(\sigma_\mu^\dagger x)=\frac{1}{2}(\sigma_\mu x)_{\nu\nu}$$

The map $\mathbb R^4\ni (x_0,x_1,x_2,x_3)\mapsto x_\mu\sigma_\mu\in V$ is an isomorphism. So V is isomorphic to $\mathbb R^4$.

Last edited: Sep 2, 2010
4. Sep 5, 2010

### LAHLH

Re: Sl(2,c)

Thanks for the help.

I convinced myself of this in the end.

$$\tilde{\sigma_{\mu}}V:=\tilde{\sigma_{\mu}}(v_{\beta}\sigma^{\beta})$$
Tracing (on the 2x2 matrix indices, not 4 vec indices, which is I think what was confusing me):
$$Tr( \tilde{\sigma_{\mu}}V)=( \tilde{\sigma_{\mu}}V)_{ii}=(\tilde{\sigma_{\mu}})_{ij}(v_{\beta}\sigma^{\beta})_{ji}=v_{\beta} (\tilde{\sigma_{\mu}})_{ij}(\sigma^{\beta})_{ji}$$

Now we have that, $$\tilde{\sigma_{\mu}}=(1,-\vec{\sigma})$$ and $$\sigma^{\mu}=(1,-\vec{\sigma})$$. Therefore if $$\mu=\beta$$ : $$(\tilde{\sigma_{\mu}})_{ij}(\sigma^{\beta})_{ji}=1_{ii}=2$$ (since e.g. 1x1=1, $$\sigma_{x}\sigma_{x}=1$$, $$\sigma_{y}\sigma_{y}=1$$, $$\sigma_{z}\sigma_{z}=1$$ etc, no summation)

On the other hand if $$\mu\neq\beta$$, we end up with $$(\tilde{\sigma_{\mu}})_{ij}(\sigma^{\beta})_{ji}$$ equal to the trace on another Pauli matrix by virtue of the cyclic identity $$\sigma_{i}\sigma_{j}=i\epsilon_{ijk}\sigma_{k}$$, and the since the Pauli matrices are traceless, this trace of the product is zero.

Combining these facts:

$$Tr(\tilde{\sigma}_{\mu}V)=v_{\beta}\delta^{\beta}_{\mu}2$$ which implies $$v_{\beta}=\tfrac{1}{2}Tr(\tilde{\sigma}_{\mu}V)$$